CHEMISTRY – Paper 1 Multiple Choice – mark scheme

B: hydrogen is a neutral gas that reacts with neither the basic drying agent (calcium oxide) nor the acidic one (concentrated sulfuric acid). A: ammonia is a basic gas and reacts with concentrated sulfuric acid, so it can be dried with calcium oxide only. C, D: hydrogen chloride and sulfur dioxide are acidic gases and react with calcium oxide, a basic oxide; they can be dried with concentrated sulfuric acid only.

A: M has spots level with the spots of P and R, and N has spots level with P and S; only P appears in both mixtures (a dye travels the same distance whether it is on its own or in a mixture). B: Q is in neither mixture: the spots of M and N are not level with the spot of Q. C: R is in M only. D: S is in N only.

D: vegetable oil and water are immiscible liquids that form two layers, so the lower layer can be run off using a separating funnel. A: ethanol and water are miscible; they are separated by fractional distillation. B: sand is an insoluble solid in a liquid; it is separated by filtration. C: sodium chloride is dissolved in the water; it is separated by evaporation or crystallisation, or the water by distillation.

C: the temperature stays constant at 60 °C while Y freezes. Energy is still given out as attractive forces form between the particles, which move from random positions into the fixed, regular positions of a solid. A: this describes boiling (the particles gain energy and spread out), the opposite change. B: the particles of a solid still vibrate about their positions; they do not stop moving, and the average kinetic energy does not change while the temperature is constant. D: this describes melting, the reverse of the change in region Q.

D: both atoms have 17 protons, so a neutral atom of each has 17 electrons; the chemical properties depend on the electrons, so the isotopes react in the same way. A: the nucleon numbers are different (35 and 37) and the numbers of protons are the same. B: both have 17 protons, so both are chlorine; isotopes differ only in the number of neutrons. C: the numbers of electrons are the same (17); the different mass does not change the number of electrons.

B: Q has 2 outer electrons and loses them to form Q²⁺; R has 5 outer electrons and gains 3 to form R³⁻, a full outer shell of 8. The charges balance when there are 3 Q²⁺ (6+) and 2 R³⁻ (6−): Q₃R₂. A: the ion charges are crossed over the wrong way round (Q₂R₃). C, D: the numbers of outer electrons (2 and 5) are used as the subscripts, either directly (C) or crossed over (D), instead of the charges on the ions.

C: each oxygen atom has 6 outer shell electrons and shares 2 of them in the double bond (O=O), leaving 4 non-bonding outer shell electrons (2 lone pairs) on each atom, i.e. 8 per molecule. Number of molecules = 0.10 × 6.02 × 10²³ = 6.02 × 10²²; non-bonding outer shell electrons = 8 × 6.02 × 10²² = 4.8 × 10²³. A: counts the electrons in one shared pair (2 per molecule). B: counts the 4 electrons shared in the double bond (4 per molecule), not the non-bonding ones. D: counts all 12 outer shell electrons of the two atoms (6 per atom), including the 4 that are shared.

A: in diamond every carbon atom forms four strong covalent bonds to its neighbours throughout the structure; this makes diamond very hard, so it can cut other materials. B: this describes graphite, whose layers slide over each other (soft, a lubricant). C: this describes the bonding in a metal; diamond is a non-metal. D: this describes an ionic compound; diamond contains only carbon atoms, joined by covalent bonds.

D: the larger tin atoms disturb the regular layers of copper atoms, so the layers cannot slide over each other as easily when a force is applied; the alloy is harder than the pure metal. A: an alloy is a mixture, and ionic bonds are not present between metal atoms. B: an alloy is a mixture of the elements, not a compound; no covalent bonds form between the atoms. C: the hardness is not caused by extra delocalised electrons; it is caused by atoms of different sizes disrupting the layers.

B: one Pb²⁺ ion combines with two I⁻ ions to form the insoluble solid PbI₂; atoms and charges balance (2+ and 2×(1−)) and the precipitate is (s). A: PbI is not the formula of lead(II) iodide; the charges do not balance (2+ and 1− on the left, no charge on the right). C: balanced, but lead(II) iodide is a precipitate, so its state symbol should be (s), not (aq). D: this is the full equation with formulae of compounds, not an ionic equation (the spectator ions K⁺ and NO₃⁻ are not removed).

C: n(Al) = 1.35 ÷ 27 = 0.050 mol; n(HCl) = 0.100 × 1.0 = 0.100 mol. 0.050 mol of Al needs 0.150 mol of HCl, so the acid is the limiting reactant. n(H₂) = 0.100 × 3/6 = 0.050 mol; volume = 0.050 × 24 = 1.2 dm³. A: takes the ratio Al : H₂ as 2 : 1 (0.025 mol) instead of 2 : 3. B: inverts the ratio (0.050 × 2/3 = 0.033 mol). D: uses the aluminium (0.050 × 3/2 = 0.075 mol) and ignores that the aluminium is in excess.

B: M_r of CuCO₃ = 64 + 12 + 48 = 124, so n = 6.2 ÷ 124 = 0.050 mol. The ratio CuCO₃ : CuO is 1 : 1, so n(CuO) = 0.050 mol; M_r of CuO = 64 + 16 = 80; mass = 0.050 × 80 = 4.0 g. A: halves the amount (a 2 : 1 ratio is assumed). C: assumes the mass does not change (the carbon dioxide formed leaves as a gas). D: doubles the amount (a 1 : 2 ratio is assumed).

D: n(Pb) = 6.21 ÷ 207 = 0.0300 mol; n(O) = 0.64 ÷ 16 = 0.0400 mol. The ratio Pb : O = 0.030 : 0.040 = 3 : 4, so the formula is Pb₃O₄. A: rounds the ratio 1 : 1.33 down to 1 : 1. B: rounds 1 : 1.33 up to 1 : 2. C: 2 : 3 is only close to 3 : 4 (0.67 against 0.75); the ratio must be multiplied up to whole numbers (× 3), not rounded.

C: n(H₂SO₄) = 0.0250 × 0.040 = 0.0010 mol; the ratio NH₃ : H₂SO₄ is 2 : 1, so n(NH₃) = 0.0020 mol; concentration = 0.0020 ÷ 0.0200 = 0.10 mol / dm³; × 17 = 1.7 g / dm³. A: uses a 1 : 1 ratio (0.0010 ÷ 0.0200 = 0.050 mol / dm³). B: divides by the volume of the acid (25.0 cm³) instead of the volume of the ammonia (0.0020 ÷ 0.0250 = 0.080 mol / dm³). D: applies the 2 : 1 ratio twice (0.0040 mol).

B: methanoic acid is only partially ionised, so its solution has fewer ions and a lower electrical conductivity. But the two solutions contain the same amount of acid (same volume and concentration); as H⁺ ions are used up, more molecules ionise, so the calcium carbonate (in excess) reacts with all of the acid and the same total volume of carbon dioxide forms. A, C: students wrongly assume a weak acid gives a smaller volume of gas, as if only the ionised fraction could react. C, D: a weak acid is said to conduct better (the conductivity of a weaker acid is lower, because there are fewer ions).

A: at pH 7 the solution is neutral: the concentration of H⁺ ions equals that of OH⁻ ions (the acid has just been neutralised). B: the start of the experiment, pH 1: the acid is at its highest concentration of H⁺ ions. C: pH about 2: still acidic, with H⁺ ions in excess because not enough alkali has been added. D: pH 13: alkaline, with OH⁻ ions in excess because too much alkali has been added.

A: the possible products, ammonium chloride and potassium carbonate, are both soluble (ammonium salts and the salts of Group 1 metals are soluble), so no precipitate forms. B: barium sulfate is insoluble. C: calcium carbonate is insoluble (carbonates are insoluble except those of Group 1 and ammonium). D: silver chloride is insoluble (chlorides are soluble except those of silver and lead).

C: zinc carbonate is insoluble, so when some remains undissolved all of the acid must have reacted; the unreacted solid is removed by filtration, leaving a solution of zinc chloride that is not contaminated with acid. A: the purpose of the excess is to make sure all the acid has reacted, not to speed up the reaction. B: the solution is made saturated later by heating off some of the water. D: carbon dioxide leaves the mixture as bubbles; the amount of solid has no effect on this.

A: a light blue precipitate that is insoluble in excess aqueous sodium hydroxide but dissolves in excess aqueous ammonia to give a dark blue solution shows Cu²⁺; ammonia produced when the solution is warmed with sodium hydroxide and aluminium foil shows nitrate. B: copper(II) is right, but a sulfate would not give ammonia in test 3 (it gives a white precipitate with acidified barium nitrate). C, D: zinc hydroxide is white, not light blue (and dissolves in excess sodium hydroxide); D also has the wrong anion.

D: sulfur dioxide is a reducing agent and decolourises acidified potassium manganate(VII); carbon dioxide does not. A: both gases give a white precipitate with limewater (calcium sulfite and calcium carbonate), so this does not distinguish them. B: both are acidic gases and turn damp blue litmus paper red. C: sulfur dioxide does not support combustion either, so a lighted splint is also put out in it.

C: magnesium is oxidised (oxidation state 0 to +2, it loses electrons) and the carbon in carbon dioxide is reduced (+4 to 0): the oxidation states change, so it is redox. A: thermal decomposition, with Ca +2, C +4 and O −2 throughout, so no oxidation state changes. B: a precipitation reaction in which ions simply swap partners, with no change in oxidation state. D: neutralisation (an acid and a base), again with no change in any oxidation state.

A: the brown colour is iodine, formed when iodide ions are oxidised (2I⁻ → I₂ + 2e⁻); the iodide ions lose electrons to W, so W is the substance that takes electrons in: an oxidising agent. B: the role is right but the electron transfer is reversed (an oxidising agent accepts electrons). C: the transfer is right but it is described as a reducing agent (a reducing agent donates electrons). D: a reducing agent donates electrons, but here the iodide ions lose electrons (they are oxidised to brown iodine), so W accepts them and is the oxidising agent.

D: molten zinc iodide contains Zn²⁺ and I⁻ ions. Positive ions are attracted to the negative electrode (the cathode, P), where Zn²⁺ gains two electrons to form zinc metal (reduction); iodine forms at the positive electrode Q. A, B: zinc cannot form at the positive electrode, because Zn²⁺ ions are repelled by it. C: the electrode is right, but zinc ions are reduced (they gain electrons); a loss of electrons (oxidation) happens to the iodide ions at Q.

B: the key is the cathode, where Cu²⁺(aq) + 2e⁻ → Cu(s) coats it with copper. The copper rod is the anode; copper atoms lose electrons more easily than the water or sulfate ions, so Cu(s) → Cu²⁺(aq) + 2e⁻ and the rod dissolves, which keeps the concentration of copper(II) ions constant. A: oxygen is formed at an inert anode; here the copper electrode itself reacts. C: hydrogen does not form at the cathode because copper (below hydrogen in the reactivity series) is discharged in preference to H⁺; oxygen would form at an inert anode, but here the copper anode reacts. D: the changes are swapped between the two electrodes (copper is deposited at the negative electrode, the cathode).

C: a T³⁺ ion has lost 3 electrons, so an atom of T has 10 + 3 = 13 electrons (and 13 protons), electronic structure 2,8,3: 3 outer shell electrons → Group 13; 3 occupied shells → Period 3 (aluminium). A: the charge is used as the Group number (and the number of shells is correct). B: the Group is right, but the Period is taken from the shells of the ion (2) instead of the atom (3). D: the 3 electrons are subtracted instead of added (7 electrons in the atom, 2,5: Group 15, Period 2).

D: down Group 17 the elements get darker in colour and change from gas to liquid to solid, so astatine, below iodine, is predicted to be a dark (black) solid. A: a gas, as for chlorine at the top of the group. B: a lighter liquid, as if the colour got paler and the state stopped at liquid. C: the appearance of bromine, which is above iodine in the group.

A: a more reactive halogen displaces a less reactive halogen from its halide solution; reactivity falls down Group 17 (Cl₂ > Br₂ > I₂), so chlorine displaces bromide ions to form bromine. B: bromine is less reactive than chlorine, so it cannot displace chloride ions. C: iodine is less reactive than bromine. D: iodine is less reactive than chlorine. (B, C and D all have the displacement the wrong way round.)

D: the compound is neutral: two Fe³⁺ ions (+6) balance three SO₄²⁻ ions (−6), so Fe₂(SO₄)₃. This shows the variable oxidation state of a transition element. A: this is iron(II) sulfate (1 : 1). B: the numbers are placed the wrong way, giving one iron ion (+3) and three sulfate ions (−6). C: the charges are swapped (three Fe and two SO₄), giving +9 and −4.

B: P is the only metal whose oxide is not reduced by carbon, so it is the most reactive (it is also the one that reacts most vigorously with acid). Of the other three, S reacts rapidly with the acid, Q only slowly and R not at all, so S > Q > R. The order is P, S, Q, R. A: Q and S are swapped (a slow reaction with acid taken to mean a more reactive metal). C: the order is reversed (least reactive first). D: S is placed above P, ignoring that P is the metal whose oxide carbon cannot reduce.

A: the peak is 150 kJ / mol above the reactants (100 + 150 = 250) and the products are 50 kJ / mol below the reactants (100 − 50 = 50), so the reaction is exothermic. B: the activation energy is right, but the products are above the reactants (ΔH = +50, endothermic). C: the products are right, but the peak is only 50 kJ / mol above the reactants (activation energy 50, because the peak height is read as 150 on the axis). D: the products are right, but the peak is 100 kJ / mol above the reactants (activation energy 100).

A: a reaction is endothermic when more energy is taken in to break bonds than is released when bonds form; the larger the difference, the larger the positive enthalpy change. A (P): 900 − 700 = +200 kJ, the largest positive value, so the most endothermic. B (Q): 1200 − 1500 = −300 kJ and D (S): 600 − 850 = −250 kJ: more energy is released than taken in, so these are exothermic (negative enthalpy change). C (R): 1100 − 1000 = +100 kJ: endothermic, but with a smaller value than P.

D: curve Y falls faster (a faster rate) and ends at the same mass, so the same amount of carbon dioxide escapes and only the rate has increased. Powder has a larger surface area, so collisions are more frequent. A: larger lumps have a smaller surface area, giving a slower rate. B: 2.0 g of marble (0.020 mol, the acid still in excess) would lose twice the mass, so Y would end lower. C: a lower temperature gives a slower rate (fewer collisions have enough energy), so the curve would fall more slowly.

B: a catalyst provides an alternative route with a lower activation energy, so the dashed peak is lower; the reactants and the products are unchanged, so the start and end levels (and the enthalpy change) are the same. A: the dashed peak is higher, which would slow the reaction. C: the peak is lower, but the products are drawn at a lower energy, so the enthalpy change would be different. D: the products are lower and the peak is unchanged: a catalyst does not alter the enthalpy change, and the activation energy would need to fall.

B: hydrogen adds across the C=C bond and the carbon skeleton is unchanged, so the alkene must have the same branched carbon chain as 2-methylpropane: (CH₃)₂C=CH₂ (methylpropene). A, C: but-1-ene and but-2-ene have four carbon atoms but an unbranched chain; adding hydrogen gives butane, which is an isomer of 2-methylpropane. D: propene has only three carbon atoms and would give propane.

D: isomers have the same molecular formula but different structures; CH₃CH₂CH(OH)CH₃ (butan-2-ol) is C₄H₁₀O with the OH group on a different carbon atom. A: propan-1-ol is C₃H₈O (one fewer carbon atom). B: pentan-1-ol is C₅H₁₂O (one more carbon atom). C: HOCH₂CH₂CH₂CH₃ is butan-1-ol written from the other end: the same compound, not an isomer.

A: acidified manganate(VII) is an oxidising agent: it oxidises the alcohol to the carboxylic acid (butanoic acid) and is itself reduced, so the purple colour disappears. B: the colour change is right, but an ester forms when an alcohol reacts with a carboxylic acid, not when an alcohol is oxidised. C: the product is right, but the reagent would turn from purple to colourless, not the reverse. D: wrong colour change and wrong product.

C: balancing atoms: carbon, 6 − 2 × 2 = 2, so 2X contains 2 C; hydrogen, 12 − 2 × 6 = 0; oxygen, 6 − 2 × 1 = 4. So 2X = C₂O₄ and X = CO₂. B: CO has the right number of carbon atoms but only half the oxygen atoms needed. D: hydrogen atoms cannot be left over (all 12 are in the ethanol). A: methane contains hydrogen atoms that are not available, and the oxygen atoms would be missing.

C: the repeat unit is –CH(CH₃)–CH(CH₃)–: two backbone carbon atoms, each carrying one CH₃ group. Opening the C=C bond of CH₃CH=CHCH₃ (but-2-ene) gives exactly this unit. A: but-1-ene would give a polymer with a CH₂CH₃ side group on every second carbon atom. B: methylpropene would give two CH₃ groups on the same carbon atom. D: propene would give a repeat unit of two carbon atoms with a CH₃ group on only one of them.

B: the name has two parts: the alcohol part first (butan-1-ol → butyl) and the acid part second (ethanoic acid → ethanoate): butyl ethanoate. A: this uses butanoic acid (4 carbon atoms) instead of ethanoic acid. C: this uses methanoic acid (1 carbon atom). D: the two parts are swapped: ethyl butanoate is made from ethanol and butanoic acid (the alcohol and the acid are reversed).

C: in SO₂ sulfur is +4; in CaSO₄ it is +6, so sulfur is oxidised (and oxygen, 0 to −2, is reduced). A: calcium is +2 in both CaCO₃ and CaSO₄; its oxidation state does not change. B: oxygen goes from 0 in O₂ to −2, a decrease, so oxygen is reduced. D: the oxidation state of sulfur increases (+4 to +6); it does not decrease to +2.